a(n+2)=-1/a(n+1)=a(n)周期2
a(n+1)=[1-a(n)]/[1+a(n)] =2/[1+a(n)] -1
1+a(n+1)=2/[1+a(n)] =2/{2/[1+a(n-1)]}=1+a(n-1)
a(n+2)=a(n)周期2
a(n+1)=[a(n)-1]/[a(n)+1]=1-2/[a(n)+1]
a(n+2)=1-2/[a(n+1)+1]=1-2/[1-2/[a(n)+1]+1]=-1/a(n)
a(n+3)=[1+a(n)]/[1-a(n)]
a(n+4)=a(n)周期4
[a(n+2)]+[a(n+1)]+[a(n)]=0
[a(n+3)]+[a(n+2)]+[a(n+1)]=0
a(n+3)=a(n)周期4